Edit in response to the OP's comment:
Here are a few facts with proofs. Throughout, let $E$ be a complexification of a real Banach space $E_{\mathbb{R}}$.
Note that an everywhere defined linear operator $T$ on $E$ is real if and only if $T(E_{\mathbb{R}}) \subseteq E_{\mathbb{R}}$.
Proposition 1. Let $A: E \supseteq D(A) \to E$ be a linear operator which is real. If $z \in D(A)$, then $\overline{z}$ is also in $D(A)$ and we have $A\overline{z} = \overline{Az}$.
Proof. We may write $z$ as $z = x+iy$, where $x,y \in E_{\mathbb{R}}$. As $A$ is real, we have $x,y \in D(A)$ and $Ax$ as well as $Ay$ is in $E_{\mathbb{R}}$. Hence,
\begin{align*}
A\overline{z} = A(x-iy) = Ax - iAy = \overline{Ax+ iAy} = \overline{Az}.
\end{align*}
Note that the third equality above uses the fact that both $Ax$ and $Ay$ are elements of $E_{\mathbb{R}}$.
Proposition 2. Let $A: E \supseteq D(A) \to E$ be a linear operator which is real. If $\lambda$ is a real number in the resolvent set of $A$, then the resolvent $R(\lambda,A) := (\lambda - A)^{-1}$ is real, too.
Proof. Let $x \in E_{\mathbb{R}}$ and set $y := R(\lambda,A)x$. Then $y$ is an element of $D(A)$ and it can be written as $y = a + ib$ for unique $a,b \in E_{\mathbb{R}}$. Since $A$ is real, we conclude that $a,b \in D(A)$.
Clearly, $x = (\lambda - A)y = (\lambda-A)a + i (\lambda - A)b$. Since $A$ is a real operator and $\lambda$ is a real number, both vectors $(\lambda - A)a$ and $(\lambda - A)b$ are elements of $E_{\mathbb{R}}$. As the decomposition of each vector in $E$ into its real and imaginary part is unqie, we conclude that $x = (\lambda - A)a$. Hence, $R(\lambda,A)x = a \in E_{\mathbb{R}}$.
This shows that $R(\lambda,A)$ is indeed real.
Proposition 3. Let $A: E \supseteq D(A) \to E$ be a linear operator which is real.
(a) If $\lambda \in \mathbb{C}$ is an eigenvalue of $A$ with eigenvector $z \in E$, then $\overline{\lambda}$ is an eigenvalue of $A$ with eigenvector $\overline{z}$.
(b) If $\lambda \in \mathbb{C}$ is a spectral value of $A$, then so is $\overline{\lambda}$.
Proof. (a) According to Proposition 1 we have $\overline{z} \in D(A)$ and $A\overline{z} = \overline{Az} = \overline{\lambda z} = \overline{\lambda} \overline{z}$.
(b) Assume that $\overline{\lambda}$ is not a spectral value of $A$. Then $A$ is closed, so we only have to show that $\lambda - A$ is a bijective mapping from $D(A)$ to $E$.
Clearly $\lambda - A$ is injective, since otherwise it would follow from (a) that $\overline{\lambda} - A$ was not injective.
In order to show that $\lambda - A$ is surjective, let $z \in E$. Since $\overline{\lambda} - A$ is surjective, we can find a vector $c \in E$ such that $(\overline{\lambda} - A)c = \overline{z}$. Hence, we obtain
\begin{align*}
\overline{(\lambda - A)\overline{c}} = \overline{\lambda} c - Ac = (\overline{\lambda} - A) c = \overline{z}
\end{align*}
(where we used Proposition 1 for the first equality), so $(\lambda - A)\overline{c} = z$. This proves that $\lambda - A$ is indeed surjective.
Remark 4. Let $A: E \supseteq D(A) \to E$ be a linear operator which is real and let $\lambda$ be a complex number. Then the following assertions hold:
(a) The number $\lambda$ is an eigenvalue of $A$ if and only if $\overline{\lambda}$ is an eigenvalue of $A$.
(b) The number $\lambda$ is a spectral value of $A$ if and only if $\overline{\lambda}$ is a spectral value of $A$.
Proof. The "only if" implications are the content of Proposition 3, and the converse implications also follow from Proposition 3 by applying the proposition to the number $\overline{\lambda}$ instead of $\lambda$.
Remark 5. The arguments in the proof of Proposition 3 actually show that we have
\begin{align*}
\overline{R(\lambda,A)z} = R(\overline{\lambda},A) \overline{z}
\end{align*}
for each $\lambda$ in the resolvent set of $A$ and each $z \in E$. This can be considered as a generalisation of Proposition 2.
Theorem 6. Let $A: E \supseteq D(A) \to E$ be a linear opertor which is real and let $\sigma$ be a compact subset of the spectrum $\sigma(A)$ such that $\sigma(A) \setminus \sigma$ is closed. Let $P$ denote the spectral projection associated to $\sigma$.
If $\sigma$ is invariant under complex conjugation (i.e. $\overline{\lambda} \in \sigma$ for all $\lambda \in \sigma$), then the operator $P$ is real.
Sketch of the proof. By Proposition 3 (or Remark 4) the entire spectrum $\sigma(A)$ is conjugation invariant. Since $\sigma$ is compact and $\sigma(A) \setminus \sigma$ is closed, we can find a closed (and smooth) cycle $\gamma$ in $\mathbb{C}$ which circumvents no element of $\sigma(A) \setminus \sigma$ but each element of $\sigma $ exactly once. As $\sigma$ and $\sigma(A) \setminus \sigma$ are conjugation invariant, we can choose $\gamma$ to be conjugation invariant, too, meaning that walking along $\overline{\gamma}$ is the same as walking along $\gamma$ in the converse direction. By definition of the spectral projection we have
\begin{align*}
P = \frac{1}{2\pi i} \int_{\gamma} R(\mu,A) \, d\mu
\end{align*}
Applying this to a vector $z \in E$ and employing Remark 5 we conclude that
\begin{align*}
\overline{Pz} = - \frac{1}{2\pi i} \int_\gamma R(\overline{\mu}, A) \overline{z} \, \overline{d\mu} = \frac{1}{2\pi i} \int_\gamma R(\mu,A) \overline{z} \, d\mu = P\overline{z}.
\end{align*}
If $z \in E_{\mathbb{R}}$ this implies that $\overline{Pz} = P\overline{z} = Pz$, so $Pz \in E_{\mathbb{R}}$. Hence, $P$ is real.
Of course, one can prove many related results, but I hope that the above arguments suffice to give you an idea of how thinks usually work in the spectral theory of real operators.
